NWhat is the Apparent Weight of a Bicyclist on a Hump?

In summary, the combined mass of the bicyclist and bicycle is 62 kg. They go over a hump with a radius of 6m at a speed of 5m/s. The apparent weight at the top of the hump is equal to the normal force, which can be calculated by subtracting the centripetal force (290N) from the weight (607.6N). This results in a normal force of approximately 350.23N.
  • #1
tdusffx
58
0
A bicycle of bicyclist of combined masss 62 kg go over a hump of radius 6m at a speed of 5m/s. what is their combined appararent weight at the top of the hump?

is it just...

Fc = MAc

Fc =M(V^2/r)

Fc = 62(25/6)

Fc = 290N
 
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  • #2
You only took into account the centripetal force, there is another force acting on the system.

BTW "apparent weight" usually means that they are asking for the normal force (the reaction force that the hump applies).
 
Last edited:
  • #3
grrr, I have no idea then.
 
  • #4
Well think about it. If i stand on a hump and don't move then my weight is going to be w = mg but if i travel over the hump at some speed then the centripetal force will act in the opposite direction of gravity and make me lighter.
 
  • #5
oh, then normal force is what i solved for. didn't I?
 
  • #6
hmmm, ya, that's correct. Hold on let me check the problem again.
 
  • #7
man, I have no idea how to solve this one..
 
  • #8
you solved for the centripetal force

Sum of force = MA (a = 0 because no acceleration in y direction, or you can plug in a = v^2/r and keep left side as n - mg)

N - mg + Centripetal = 0

you already solved the centripetal force. Now plug in mg and solve for N to get "apparent weight" or normal force.
 
  • #9
hmmm...thanks man. I appreciate it.
 
  • #10
anyways...

Fn - mg + Fc = 0

Fn = mg - Fc

Fn = 607.6-258.33

Fn is roughly 350
 

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